Abstract. In this article, the author designs a teaching situation: developing
formula to calculate the distance from a point to a plane in space. In these situations,
all learning activities of students will have been planned by the teacher. The formula
to calculate the distance from a point to a plane will be created through the process
of two different student’s activities: the first is the process of determining the
distance in synthetic geometry and the second is the similarity in the formula for
calculating the distance from a point to a line in the plane (something that students
already know). In this teaching situation, students learn through their own activities
but in a manner that was part of the teacher’s plan. In the process of implementing
the scenario, from time to time teachers will need to orient students at a minimum
level in such a manner that students will find the desired formula.

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JOURNAL OF SCIENCE OF HNUE
Interdisciplinary Science, 2013, Vol. 58, No. 5, pp. 47-52
This paper is available online at
DESIGNING A TEACHING SITUATION: DEVELOPING FORMULA
TO CALCULATE THE DISTANCE FROM A POINT TO A PLANE IN SPACE
(Geometry for the 12th grade, Chapter 3, Lesson 2)
Bui Van Nghi1 and Nguyen Tien Trung2
1Faculty of Mathematics, Ha Noi National University of Education
2University of Education Publishing House
Abstract. In this article, the author designs a teaching situation: developing
formula to calculate the distance from a point to a plane in space. In these situations,
all learning activities of students will have been planned by the teacher. The formula
to calculate the distance from a point to a plane will be created through the process
of two different student’s activities: the first is the process of determining the
distance in synthetic geometry and the second is the similarity in the formula for
calculating the distance from a point to a line in the plane (something that students
already know). In this teaching situation, students learn through their own activities
but in a manner that was part of the teacher’s plan. In the process of implementing
the scenario, from time to time teachers will need to orient students at a minimum
level in such a manner that students will find the desired formula.
Keywords: Teaching situation, the distance from a point to a plane in space.
1. Introduction
When teaching mathematics, particularly geometry, "The teacher needs to create
situations in which students must understand the problems, do the work needed to solve
problems, adjust his thinking, and attempt to obtain new information.” [2; 93].
In each teaching situation, we believe that the teacher needs to design a structure
that contains three basic situations: situation of action, situations of comunication and
situations of validation [4].
From the point of view that “doing mathematics properly implies that one is dealing
with problems" [5; 22], we need to design a teaching situation where students are given
face-to-face situations, and they work on their own and together to solve the problem. At
that point students must adjust their thinking to the new information obtained and establish
or develop their skills.
Received November 05, 2012. Accepted June 25, 2013.
Contact Nguyen Tien Trung, e-mail address: trungnt@hnue.edu.vn
47
Bui Van Nghi and Nguyen Tien Trung
Therefore, the teaching situation must be designed in such a way that students "will
be responsible for the relationship between them and knowledge.” [6; 159]
According to a study presented by Bui Van Nghi (2008) [1; 184], in the process of
teaching analytic geometry, we "need to pay attention on both the axiomatic method and
the method of coordinates." The two methods complement each other, contributing to the
improvement of the quality of teaching geometry and the ability to learn it.
From theories presented in research, textbooks and teacher’s books, we believe
that it is feasible to design geometric teaching situations that are based upon opinions
of activities and ideas of the theory of situations.
2. Content
In this paper, we present the results of our study: designing the teaching situation to
create formula for calculating the distance from a point to a plane in space.
The scenario of the teaching situation involves the following actions:
* Action 1 (situation of action)
The teacher divides the class into four groups and asks each group to solve the
following problem: “In space, there is a plane (α) : Ax+By+Cz+D = 0 and there is a
pointM (x0; y0; z0). Let’s determine the formula of calculating the distance from a point
M to a plane (α).
Every student knows how to determine the distance from a point to a plane in space:
the distance from a pointM to a plane (α) is equal to the length of segmentMM ′ where
M ′ is the perpendicular projection of the pointM in plane (α). Then, students can propose
the process (basic process) like this:
Step 1: Determine the pointM ′ which is the perpendicular projection of pointM in
plane (α);
Step 2: Determine the length of segmentMM ′.
So, at the moment, all students have the tools: They have gone through the process
of calculating this distance previously. However, this process gives them only the segment
length (the length of segmentMM ′) and not a formula to determine the length. They have
the qualitative methods but not the quantitative method (formula) to arrive at what they
are looking for, with hope and faith in the results.
Thus, the problem of calculating the distance becomes a problem of calculating
the length of a segment or the distance between two points. This kind of problem, for
students, is a known problem (in analytic geometry, they have got the formula to solve
this problem). So, every student believes that they can do this problem.
After a discussion time, each group or individual student can be asked to perform
the tasks in the following ways:
* Action 2 (situation of action)
Option 1: (Use the base process combined with what one knows of vectors in
space). Let M ′ (x1; y1; z1) be the perpendicular projection of point M on plane (α), we
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Designing a teaching situation: developing formula to calculate the distance...
have
−−−→
MM ′⊥ (α)⇔ −−−→MM ′//~n(α), we have
−−−→
MM ′ = t~n(α), (t ∈ R) (2.1)
Moreover, we have
−−−→
MM ′ = (x1 − x0; y1 − y0; z1 − z0) , ~n(α) = (A;B;C) so the
equation (2.1) become this system of equations
x1 − x0 = tA
y1 − y0 = tB
z1 − z0 = tC
⇔
x1 = x0 + tA
y1 = y0 + tB
z1 = z0 + tC
(2.2)
Applying the coordinates of pointM ′ (in the equation (2.2)) in the equation of plane
(α), we have
A (x0 + tA) +B (y0 + tB) + C (z0 + tC) +D = 0
⇔ t = −Ax0 +By0 + Cz0 +D
A2 +B2 + C2
Then, we have
d(M ; (α)) =
∣∣∣−−−→MM ′
∣∣∣
=
√
(At)2 + (Bt)2 + (Ct)2
=
√
(A2 +B2 + C2) t2
= |t|
√
A2 +B2 + C2
Applying the value of t in the above formula, we have
d (M ; (α)) =
∣∣∣∣−Ax0 +By0 + Cz0 +DA2 +B2 + C2
∣∣∣∣ .
√
A2 +B2 + C2
=
|Ax0 +By0 + Cz0 +D|√
A2 +B2 + C2
Option 2: (Use the base process combined with knowledge of the vector in space
which is presented in the textbook). LetM ′ (x1; y1; z1) be the perpendicular projection
of point M on plane (α). We have two vectors
−−−→
MM ′ = (x1 − x0; y1 − y0; z1 − z0) and
~n(α) = (A,B,C) which are parallel to each other because they are perpendicular to plane
(α). So, we have
49
Bui Van Nghi and Nguyen Tien Trung
∣∣∣−−−→MM ′
∣∣∣ . ∣∣~n(α)∣∣ =
∣∣∣−−−→MM ′.~n(α)
∣∣∣ = |A (x− x0) +B (y − y0) + C (z − z0)|
= |Ax0 + By0 + Cz0 + (−Ax−By − Cz)| (2.3)
On the other hand, pointM ′ is in plane (α) so we have
Ax1 +By1 + Cz1 +D = 0⇔ D = − (Ax1 +By1 + Cz1)
Applying the formula in (1) we have
∣∣∣−−−→MM ′∣∣∣ . ∣∣~n(α)∣∣ = |Ax0 +By0 + Cz0 +D|
.
Having d (M, (α)) denote the distance from pointM to plane (α), we have
d(M, (α)) =
∣∣∣−−−→MM ′
∣∣∣
=
|Ax0 +By0 + Cz0 +D|∣∣~n(α)∣∣
=
|Ax0 +By0 + Cz0 +D|√
A2 +B2 + C2
Option 3: (Suggest that students solve the problem looking at the similarity in the
formula for calculating the distance from a point to a line in the plane). In the class,
students can devise a formula to calculate the distance from a point to a line in the plane.
This is therefore a formula to calculate the distance from a point to a plane in space.
Suppose that, in a similar way, it has the form
d(M, (α)) =
|Ax0 +By0 + Cz0 +D|√
A2 +B2 + C2
(2.4)
In this case, one can think quickly. The teacher needs to lead his students to test and
confirm that which is thought to be true or false. We can do this by the following:
50
Designing a teaching situation: developing formula to calculate the distance...
Let M ′ (x1; y1; z1) be the perpendicular projection of point M onto plane (α). We
can turn the formula (2.4) into the formula
MM ′ =
|Ax0 +By0 + Cz0 +D|∣∣~n(α)∣∣
⇔MM ′. ∣∣~n(α)∣∣ = |Ax0 +By0 + Cz0 +D|
⇔
∣∣∣−−−→MM ′
∣∣∣ . ∣∣~n(α)∣∣ = |Ax0 +By0 + Cz0 +D|
The angle of the two vectors
−−−→
MM ′ = (x1 − x0; y1 − y0; z1 − z0) and
~n(α) = (A,B,C) equals 0
0 or equals 1800. So, we have
∣∣∣−−−→MM ′.~n(α)
∣∣∣ =
∣∣∣
∣∣∣−−−→MM ′
∣∣∣ . ∣∣~n(α)∣∣ cos
(−−−→
MM ′, ~n(α)
)∣∣∣
=
∣∣∣−−−→MM ′∣∣∣ . ∣∣~n(α)∣∣
Inferred
∣∣∣−−−→MM ′
∣∣∣ . ∣∣~n(α)∣∣ =
∣∣∣−−−→MM ′.~n(α)
∣∣∣
= |A (x1 − x0) +B (y1 − y0) + C (z1 − z0)|
= |Ax0 +By0 + Cz0 + (−Ax1 − By1 − Cz1)| (2.5)
On the other hand, pointM ′ belongs to plane (α) and we have
Ax1 +By1 + Cz1 +D = 0⇔ D = − (Ax1 +By1 + Cz1)
Applying the above formula (2.5), we have
∣∣∣−−−→MM ′
∣∣∣ . ∣∣~n(α)∣∣ = |Ax0 +By0 + Cz0 +D|
.
So, the previous judgement is seen to be correct.
* Action 3: Validation and applying the formula to exercises (situation of action
and validation)
Exercise 1. Calculate the distance from point to plane (α) in space for each of the
following cases:
a)M (1;−2; 13) , (α) : 2x− 2y − z + 3 = 0
b)M (1;−2; 3) , (α) : x− 5y − 2z + 5 = 0
c)M (1; 2; 2) , (α) : 3x− y − z + 1 = 0
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Bui Van Nghi and Nguyen Tien Trung
d)M (−3; 1; 5) , (α) : z = 0
Exercise 2. Calculate the distance between two parallel planes (P ), (Q):
(P ) : x+ 2y + 2z + 11 = 0; (Q) : x+ 2y + 2z + 2 = 0
Exercise 3. a) Let the two planes (α) : x−2y+2z−11 = 0 and (β) be two parallel
planes. Find the equation of plane (β) if the distance between planes (α) and (β) equals 3.
b) Find pointM which is on line Oz such that the two planes having the equations:
x+ y − z + 1 = 0; x− y + z + 5 = 0 are equidistant.
* Action 4: Legitimization and institutionalization (situation of validation)
The teacher leads his students to come to the conclusion: In space Oxyz, the
distance from point M (x0; y0; z0) to the plane (α) : Ax + By + Cz + D = 0 equals
d(M, (α)) =
|Ax0 +By0 + Cz0 +D|√
A2 +B2 + C2
.
The teacher also needs to let his students experience this process: when they face
cognitive obstacles, they can look at how they can simplify it and look for aspects that are
already known. So, the two formulas of distance (the distance from a point to a line in the
plane and the distance from a point to a plane in space) have similar forms.
3. Conclusion
In the teaching scenario presented above, students play a role in the discovery in an
encouraging and interactive way. In the process of implementing such a scenario, teachers
can periodically provide support to orient students at a minimum level to allow students
to find the desired formula.
REFERENCES
[1] Bui Van Nghi, 2008. Methodology of teaching Mathematics in detail. University of
Education Publishing House, Hanoi.
[2] Bui Van Nghi, 2008. Applying theory of teaching to teach Mathematics. University
of Education Publishing House, Hanoi.
[3] Geometry Text books for the 12th Grade, Teacher’s books, Geometry Student
Exercise books for the 12th Grade (basic and advance), Vietnam Education
Publishing House, Hanoi.
[4] Virginia M. Warfield, 2006. Invitation to Didactique. University of Washington,
Seattle, Washington.
[5] Guy Brousseau, 2002. Theory of Didactical situations in mathematics. Vol. 19,
Kluwer Academic Publishers.
[6] Annie Bessot, Claude Comiti, Le Thi Hoai Chau, Le Van Tien, 2009. Elements
of Theory of Didactical situations in Mathematics. Bilingual book Vietnamese -
French, Ho Chi Minh National University of Education Publishing House.
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